2004
DOI: 10.4134/bkms.2004.41.2.359
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Radicals of a Left-Symmetric Algebra on a Nilpotent Lie Group

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Cited by 4 publications
(7 citation statements)
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“…If we believe that rad(A) is an ideal in this case, it follows that rad(A) ⊆ nil(A) and hence rad(A) = nil(A). More generally the following result is proved in [22]:…”
Section: Its Radical Is Given Bymentioning
confidence: 95%
See 1 more Smart Citation
“…If we believe that rad(A) is an ideal in this case, it follows that rad(A) ⊆ nil(A) and hence rad(A) = nil(A). More generally the following result is proved in [22]:…”
Section: Its Radical Is Given Bymentioning
confidence: 95%
“…Also it need not coincide with the Koszul radical rad(A) of A. But we have the following result [21]: Suppose that the Lie algebra g A is nilpotent. Since rad(A) is complete the Lemma implies that rad(A) is left-nilpotent.…”
Section: Its Radical Is Given Bymentioning
confidence: 97%
“…Now we shall briefly discuss the problem of extension of a left-symmetric algebra by another left-symmetric algebra. To our knowledge, the notion of extensions of left-symmetric algebras has been considered for the first time in [9], to which we refer the reader for more details [15].…”
Section: Recall That a Lie Algebra mentioning
confidence: 99%
“…To our knowledge, the notion of extensions of left-symmetric algebras has been considered for the first time in [9], to which we refer the reader for more details. See also [4].…”
Section: Left-symmetric Algebras and Their Extensionsmentioning
confidence: 99%